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JulsSmile [24]
3 years ago
13

3C+4D=5 2C+5D=2 solve for d and c

Physics
1 answer:
lara31 [8.8K]3 years ago
7 0

Answer:

C=\frac{17}{7}, D=-\frac{4}{7}

Explanation:

The system of equation is the following:

3C+4D=5\\2C+5D=2

We can solve the system by expliciting C from the second equation. We get:

2C+5D=2\\2C=2-5D\\C=\frac{2-5D}{2}=1-\frac{5}{2}D

And if we know substitute C into the first equation, we find

3C+4D=5\\3(1-\frac{5}{2}D)+4D=5\\3-\frac{15}{2}D+4D=5\\3-\frac{15}{2}D+\frac{8}{2}D=5\\3-\frac{7}{2}D=5\\-\frac{7}{2}D=2\\-7D=4\\D=-\frac{4}{7}

And by substituting D into the second equation, we find

C=1-\frac{5}{2}D=1-\frac{5}{2}(-\frac{4}{7})=1+\frac{10}{7}\\C=\frac{7}{7}+\frac{10}{7}=\frac{17}{7}

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4 0
4 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
If a car accelerates uniformly from rest to 15 meters
Talja [164]

Answer:

1.125m/s^2

Explanation:

Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically

v^2= u^2+2as

Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.

a = ?

u = 0m/s

v = 15m/s

s = 100m

Substituting the values into the formula above

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225= 200a

Divide both sides by 200

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a= 1.125m/s^2

Hence the acceleration of the car is 1.125m/s^2.

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